144 Homework Solutions for Engineering Newnan, '-- Chapter 6: Annual Economic Eschenbach, Cash Analysis, 9th Edition Lavelle Flow Analysis 1\ ~;),11~l 6 -rn" 1 f~:~ie'~J ,.,c,' $60 ~':1 $45 $30 $15 ;; c c c : ,- c , "" C = $15 + $15 (AlG, 10%,4) = $15 + $15 (1.381) = ~35.72 6 ! ..":::'_1.._L.. -' 2 '-:la, i~::)I B B B B B j " ,,~ $100 $100 B = [$100 + $100 (F/P, 15%,4)] (AlF, 15%,5) = [$100 + $100 (1.749)] (0.1483) = ~40.77 6-3 I I $60 $45 E E .$30 i j j E E --.' E = $60 -$15 (AlG, 12%,4) = $60 -$15 (1.359) = $39.62 I ;i' ! ~ r ",./, Homework Solutions for Engineering Economic Analysis, 9th Edition Newnan, Eschenbach, Lavelle 6-32 Around the Lake $75,000 $3,000/ r $7,500/ r $1,500/ r $45,000 15 ears First Cost Maintenance Annual Power Loss Pro en Taxes Salva e Value Useful Life Under the Lake $125,000 $2,000/ r $2,500/ r $2,500/ r $25,000 15 ears Around the Lake EUAC = $75,000 (AlP, 7%,15) + $12,000 -$45,000 (AlF, 7%,15) = $75,000 (0.1098) + $12,000 -$45,000 (0.0398) = $18.444 Under the Lake EUAC = $125,000 (AlP, 7%,15) + $7,000 -$25,000 (AlF, 7%,15) = $125,000 (0.1098) + $7,000 -$25,000 (0.0398) = $19.730 Go around the lake. 6-33 :'~c' ~Jji 1(..i1 Engineering Department Estimate $30 : $27 ~ i ' If.; .$9 $24 Amounts x 103 $21 $] 8 $15 $12 IOE $6 EUAC = $30,000 -$3,000 (AlG, 8%, 10) = $30,000 -$3,000 (3.871) = $18,387 ;~:(r:r t:!Yro-clean'soffer of $15.000/vr is less costlv. " 'i ~ . 157 -160 Homework Solutions for Engineering Economic Analysis, 9th Edition Newnan, Eschenbach, Alternative EUAB i A -EUAC = $845 Alternative B EUAB -EUAC To I Lavelle maximize -$3,000 = $1,400 (EUAB (0.30672) -$5,000 -EUAC) = -$75.16 (0.30672) choose = -$133.60 alternative A. (less neaative value). 6-37 Machine X EUAC = $5,000 (AlP, = $5,000 (0.2505) 8%, 5) =$1,252 Machine Y EUAC = ($8,000 =$1,106 Select Machine -$2,000) (AlP, 8%,12) + $2,000 (0.08) + $150 ! .. Y. 6-38 Annual Cost of Diesel Annual Cost of Gasoline EUACdiesel Fuel = ($13,000 = [$50,000km/(35 km/i)] x $0.48/1 = $685.71 = [$50,000km/(28 km/i)] x $0.51/1 = $910.71 -$2,000) + $685.71 = $11,000 (AlP, fuel (0.2886) 6%, + $300 + $120 4) + $2,000 repairs + $500 (0.06) insurance + $1,485.71 = $4,780.31 EUACgasoline = ($12,000 -$3,000) + $910.71 (AlP, fuel 6%, + $200 3) + $3,000 repairs + $500 . (0.06) insurance = $5,157.61 The diesel taxi is more economical. 6-39 Machine A EUAC = $1,000 + Pi ~i~- = $1,000 + $10,000 (AlP, 10%,4) -$10,000 (AlF, 10%,4) =$1,000+$1,000 , =$2,000 ( ,c..'cc 7 , -- 162 Homework Solutions for EngineeringEconomic Analysis, 9thEdition Newnan, Eschenbach, Lavelle 6-42 I Seven year analysis period: Alternative A EUAB -EUAC = $55 -[$100 + $100 (P/F, 10%,3) + $100 (P/F, 10%,6)] (AlP, 10%,7) = $55 -[$100 + $100 (0.7513) + $100 (0.5645)] (0.2054) = +$7.43 Alternative B EUAB -EUAC = $61 -[$150 + $150 (P/F, 10%,4)] (AlP, 10%,7) ": $61 -[$150 + $150 (0.683)] (0.2054) =~~ .Q1!QQ'§~ . Note: The analysis periOd is seven years, hence one cannot compare three §: years of A vs. four years of B, If one does, the problem is Constructed so he will get the wrong answer. 6-4S 6-43 Us EUACgas ": (P -S) (AlP, 1'10,n) + SL + Annual Costs = ($2,400 -$300) (AlP, 10%,5) + $300 (0.10) + $1,200 + $300 = $2,100 (0.2638) + $30 + $1,500 = $2,084 -.. He: Np' EUACelectr= ($6,000 -$600) (AlP, 10%, 10) + $600 (0.10) + $750 + $50 = $5,400 (0.1627) + $60 + $800 = $1,739 .. c NPli\ '§~l~!Jb~m~mQ-mQ!Qr .~ 6-44 .QhQQ EUAC Comparison EqUlv Gravity Plan Initial Investment: = $2.8 million (AlP, 10%, 40) = $2.8 million (0.1023) Annual Operation and maintenance Annual Cost EUAC/ , I "' .= $286,400 ~\ ' = $10,000 = $296,400 E UACa ~ .QbQ~ ) Homework Solutions for Engineering Economic Analysis, 9th Edition 163 Newnan, Eschenbach, Lavelle Pumping Plan Initial Investment: = $1.4 million (AlP, 10%, 20) = $1.4 million (0.1023) Additional investment in 10th year: = $200,000 (P/F, 10%, 10) (AlP, 10%, 40) = $200,000 (0.3855) (0.1023) = $7,890 Annual Operation and maintenance = $25,000 Power Cost: = $50,000 for 40 years = $50,000 Additional Power Cost in last 30 years: = $50,000 (F/A, 10%,30) (AlF, 10%,40) = $50,000 (164.494) (0.00226) = $18,590 Annual Cost = $244,680 Select the Pumoing Plan." = $143,200 . 6-45 Use 20 year analysis period. I Net Present Worth Approach ~ NPWMas. = -$250 -($250 -$10) [(P/F, 6%,4) + (P/F, 6%, 8) + (P/F, 6%, 12) + (P/F, 6%,16)] + $10 (P/F, 6%, 20) -$20 (PIA, 6%, 20) = -$250 -$240 [0.7921 + 0.6274 + 0.4970 + 0.3936) + $10 (0.3118) -$20 (11.470) =-$1,031 NPWBRK= -$1,000 -$10 (PIA, 6%, 20) + $100 (P/F, 6%, 20) = -$1,000 -$10 (11.470) + $100 (0.3118) = -$1,083 ~hoose Masonite to save $52 on Present Worth of Cost. Equivalent Uniform Annual Cost Approach ~ C,c' EUACMas. = $20 + $250 (AlP, 6%, 4) -$10 (AlF, 6%, 4) = $20 + $250 (0.2886) -$10 ! (0.2286) ..I '" = $90 EUACBRK = $10 + $1,000 (AlP, 6%, 20) -$100 (AlF, 6%, 20) .;.,," = $10 + $1,000 (,0.872) -$100 (0.0272)" = $94 ""! (1:1;, ;~.,)i "...I!J .QhQose Masonate to save $4 oer ~ear. d 1- ,.r ~